Question

Health Canada wishes to estimate the mean amount of mercury μ, in parts per million (ppm), in commercial fish. Specifically, they want to estimate the value of μ to within 0.03, and be 98% confident in their results. Assume that the amount of mercury in commercial fish varies from one fish to another, the standard deviation being σ=0.4 ppm. How many commercial fish should be randomly chosen, each chosen will have the mercury content measured, to satisfy Health Canadas estimation of μ?

Answer 1

Provided that a desired margin of error $E$ is given, the population standard deviation $\sigma$ is provided, and the significance level is specified, we can compute the minimum required sample size that will lead to a margin of error less than or equal to the one specified, by using the following formula:

$$n\geq({z_c\sigma \over E})^2$$

The critical value for $\alpha=0.02$ is $z_c=z_{1-\alpha/2}=2.326$

Given that $E=0.03, \sigma=0.4.$ Then

$$n\geq({2.326\cdot0.4 \over 0.03})^2$$

$$n\geq962$$